Sass, same properties on diffrent class - css

I have some code, which looks sth like this:
.elephant
background: yellow
.young-eph
color: white
.tiger
background: yellow
.young-tgr
color: white
How can I improve this code?
In css you can write same properties for multiple classes with comma
.elephant, .tiger {
background: yellow
}
but how to do that with sass?

You pretty much do the same thin as in normal css. Just put a comma in-between.
.elephant, .tiger {
background: yellow
.young-eph, .young-tgr {
color: white
}
}
In this case I'd replace .young-eph and .young-tgr by .young tho. having 2 different classes in this case looks redundant.
.elephant, .tiger {
background: yellow
.young {
color: white
}
}
EDIT after new info
If you have deep inheritance like that I'd just change the classes a bit. I generally avoid using extend.
if you use the classes .eph, .tgr, .adult, .young and .very
you can pretty much just do this:
.adult {
background: yellow
.young {
color: white
&.very {
padding-left: 30px;
}
}
}
& means the object has both .young and .very, so depending on needs you might not need it
you also most likely don't have to nest all of them, more code (including some html) might be nice.

The syntax is the same. Sass is transpiled to css, which means you can write good ole css in your .scss-files as well as sass.
This is valid sass:
.elephant, .tiger {
background: yellow;
}
.elephant {
border: dotted blue 1px;
.young-eph {
color: cyan;
}
}
.tiger {
border: dotted red 1px;
.young-tgr {
color: magenta;
}
}
Here's a link to a plunker.
EDIT: To your next question:
.elephant
background: yellow
.young-eph
color: white
.very-young-eph
padding-left: 30px
Maybe You can make a class .animal and inside [class*="young-"] and [class*="very-young-"] which you can then use to extend.
.animal
background: yellow
[class*="young-"] {
color: white
[class*="very-young-"] {
padding-left: 30px

Related

Getting the default state color (currentColor) in a different state such as hovered

Is there a keyword like currentcolor which allows us to get the color of a class in its default state?
For example, I'm trying to create a re-useable button style, and currentcolor keyword helps a lot until I try to create the :hovered state.
.outline-btn {
background-color: transparent;
border: 1px solid currentColor;
padding: 0.5em 1.5em;
}
.rounded-btn {
border-radius: 50px;
}
The default state looks the way we want and changing the color or the font-size would also adjust the rest of the properties.
But we want the :hovered state to invert the colors (white text and orange background in this case)
.outline-btn:hover, .outline-btn:active, .outline-btn:focus {
background-color: currentcolor;
color: white;
}
But since in this state the color becomes white, everything else also turns white.
Is there a way that we can achieve the behavior that we want without having to create multiple classes for the different button styles that we want?
Desired effect on hover:
Also I forgot to mention that I am using SCSS if that helps.
Thanks for your time :)
If you think about it, you're essentially wanting currentColor to act as a variable -- to hold a constant value. The upcoming CSS variables will help with this, but until they're better supported, we have Sass variables.
By defining the colors as variables you can write them out very verbosely and specifically, but only have to change the color in one place when needed.
$btn-color: red;
$btn-bg: transparent;
.outline-btn {
background-color: $btn-bg;
border: 1px solid $btn-color;
padding: 0.5em 1.5em;
color: $btn-outline-color;
&:hover,
&:active,
&:focus {
background-color: $btn-outline-color;
color: $btn-outline-bg;
}
}
You could go a step further and have those variables set to equal previously set variables you're using for the body/html color background, e.g., $bg-bg: $body-bg; $btn-color: $text-color;. I love currentColor as well and this isn't as clean as that, but it might be more appropriate in this case.
You can then build this out as a mixin as user6292372 noted. Something like:
#mixin buttonBuilder($color, $bg) {
background-color: $bg;
border: 1px solid $color;
color: $color;
&:hover {
background-color: $color;
color: $bg;
}
}
...
.outline-btn {
#include button-builder($btn-color, $btn-bg);
}
Then you can easily make multiple variants.
this can't be done with css only
if you use helpers like SCSS or Less you could make yourself a mixin where you only insert the color you want as a parameter.
but you would still have to make several classes (as many as you need different colors) but can reuse your mixin within like this (scss example):
#mixin invertHover($color) {
background-color: transparent;
border: 1px solid $color;
color: transparent;
&:hover {
background-color: $color;
border: 1px solid transparent;
color: $color;
}
}
.blue-box { #include invertHover('blue'); }
.black-box { #include invertHover('#000000'); }

Confused about overriding CSS styles

I understand CSS basics, but I keep running into trouble with conflicting styles. Consider the following styles.
First, the default font color in my style sheets is black. I want that color applied to all picture captions - unless they're contained in divs with a class CoolL or CoolR...
.CoolL .Caption, .CoolR .Caption { color: #900; }
Now all the captions in the Cool series have brown text. But there are situations where I want the captions to have a black background with white text, so I created this rule:
.Black { background: #000; color: #fff; }
Now consider the following HTML. Class Caption by itself should have black text. However, this is inside a div with a class CoolR, so it displays brown text instead. But I added the class Black to the last div, which should change the background to black and the text color to white...
<div class="CoolR Plus Max300">
<div class="Shadow2">
<img src="">
<div class="Caption Black">Text</div>
</div>
</div>
In fact, the background is displaying black, but the text color is still brown.
I get these problems all the time, and the only way I can fix them is to write long, detailed styles, like this...
.Black, .Caption .Black, .CoolR .Caption.Black, .EverythingElseThatCouldBeBlack .Black { background: #000; color: #fff; }
What am I missing? Thanks.
I think you are over complicating things. This will become a maintenance issue as you add more styles. I would define separate classes and keep things simple. It's also important to understand CSS specificity.
.caption {
color: #000;
}
.cool-caption {
color: #900;
}
.caption-with-background {
background-color: #000;
color: #fff;
}
You could try :
.Black { background: #000 !important; color: #fff !important; }
There are a few fixes, but as previously recommended you should mark all of the settings you want to override previous ones with !important. With that, your code would look like this:
.Black {
background: #000;
color: #fff;
}
Also, not sure if you asked this, but you can apply CSS to all components by using the *, like so:
* {
//blahblahblah
}
you are defining the first case with a descendant selector which overrides the second class, which is merely a class. every answer given already will work but are entirely unnecessary. just add this to your style sheet:
.CoolR1 .Black, .Black{ background: #000; color: #fff;}
/** you could also chain your classes for specificity power **/
.Black.Caption{color:#fff}
that should do it. you can read more about selectors here:
http://docs.webplatform.org/wiki/css/selectors
I think that generally a more specific rule overrides a more general one, thus the more specific '.CoolR .Caption' is overriding the more general .Black. You'll probably be able to override this with !important, but a better style might be to reduce the complexity of your rules:
.Cool .caption { color: #900; }
.Cool .caption.black { color: background: #000; color: #fff; }
And put .L and .R in separate classes
.Cool.L { . . . } /* For things specific to CoolL, but not CoolR */
.Cool.R { . . . } /* and vice-versa */

Set a variable in Sass depending on the selector

I’ve got a website that’s using a few different ‘main’ colors. The general HTML layout stays the same, only the colors change depending on the content.
I was wondering if I could set a color variable depending on the CSS selector. This way I can theme my website with a few variables and let Sass fill in the colors.
For example:
$color-1: #444;
$color-2: #555;
$color-3: #666;
$color-4: #777;
body.class-1 {
color-default: $color-1;
color-main: $color-2;
}
body.class-2 {
color-default: $color-3;
color-main: $color-4;
}
/* content CSS */
.content {
background: $color-default;
color: $color-main;
}
I was thinking of using a mixin for this, but I was wondering if there’s a better way to do this—with a function maybe? I’m not that great with Sass, so any help would be appreciated.
I think a mixin is the answer. (As I wrote, variables won’t work.)
#mixin content($color-default, $color-main) {
background: $color-default;
color: $color-main;
}
body.class-1 {
#include content(#444, #555);
}
body.class-2 {
#include content(#666, #777);
}
That SCSS compiles to this CSS:
body.class-1 {
background: #444444;
color: #555555; }
body.class-2 {
background: #666666;
color: #777777; }
If you wanted to group the color values together in your SCSS file, you could use variables in conjunction with the mixin:
$color-1: #444;
$color-2: #555;
$color-3: #666;
$color-4: #777;
body.class-1 {
#include content($color-1, $color-2);
}
body.class-2 {
#include content($color-3, $color-4);
}
as sass documentation explain nicely (https://sass-lang.com/documentation/variables):
Sass variables are all compiled away by Sass. CSS variables are included in the CSS output.
CSS variables can have different values for different elements, but Sass variables only have one value at a time.
Sass variables are imperative, which means if you use a variable and then change its value, the earlier use will stay the same. CSS variables are declarative, which means if you change the value, it’ll affect both earlier uses and later uses.
We may take advantage of that using a combination of sass and css variables to achieve what you want:
//theme colors
$red-cosmo: #e01019;
$green-cosmo: #00c398;
$primary-color: var(--primary-color);
body{
--primary-color: #{$red-cosmo};
}
body.univers-ride{
--primary-color: #{$green-cosmo};
}
So when I call my sass variable $primary-color, it will print as my css variable "var(--primary-color)" that will expand as $green-cosmo only if my body has the "univers-ride" class else it will be $red-cosmo the default color.
If you really want to get hacky you could also define your different color schemes in a single variable like $scheme1: class1 #333 #444, where the first value is always the name, and that is followed by all the colors in that scheme.
You can then use #each:
// Define your schemes with a name and colors
$scheme1: class1 #444 #555;
$scheme2: class2 #666 #777;
$scheme3: class4 #888 #999;
// Here are your color schemes
$schemes: $scheme1 $scheme2 $scheme3;
#each $scheme in $schemes {
// Here are the rules specific to the colors in the theme
body.#{nth($scheme, 1)} .content {
background-color: nth($scheme, 2);
color: nth($scheme, 3);
}
}
This will compile to:
body.class1 .content {
background-color: #444444;
color: #555555; }
body.class2 .content {
background-color: #666666;
color: #777777; }
body.class4 .content {
background-color: #888888;
color: #999999; }
Obviously if you don't want to combine body.class1 and .content in your selectors, you could just specify a mixin content($main, $default) and call it inside the #each using nth just like in the above code, but the point is you don't have to write out a rule for each of your classes.
EDIT There are lots of interesting answers on Creating or referencing variables dynamically in Sass and Merge string and variable to a variable with SASS.
You can also create a mixing that use the ampersand parent selector. http://codepen.io/juhov/pen/gbmbWJ
#mixin color {
body.blue & {
background: blue;
}
body.yellow & {
background: yellow;
}
}
UPDATE: its 2017 and variables does works!
#mixin word_font($page) {
#font-face {
font-family: p#{$page};
src: url('../../static/fonts/ttf/#{$page}.ttf') format('truetype');
font-weight: normal;
font-style: normal;
}
.p#{$page} {
font-family: p#{$page};
}
}
// Loop and define css classes
#for $i from 1 through 604 {
#include word_font($i);
}
If you don't want to use a variable for each color, you can use one variable for all kinds of colors. In the mixin you can choose the right color with nth. For instance, if you write the index of the color as 1, then you get the first color in the color variable.
$colors: #444, #555, #666, #777;
#mixin content($color-default-num, $color-main-num) {
background: nth($colors, $color-default-num);
color: nth($colors, $color-main-num);
}
body.class-1 {
#include content(1, 2);
}
For me the definite answer to my problem was creating a map of maps and loopig through them as follows:
$pallettes: (
light-theme: (
container-color: red,
inner-color: blue,
),
dark-theme: (
container-color: black,
inner-color: gray,
),
);
#each $pallette, $content in $pallettes {
.main.#{$pallette} {
background-color: map-get($content, container-color);
.inner-div {
background-color: map-get($content, inner-color);
}
}
}
You can simply override your scss variables inside of the class wrapper:
$color1: red;
$color2: yellow;
header { background: $color1; }
.override-class {
$color1: green;
header { background: $color1; }
}
Seems to work for me.

SASS checking against a class name

Hi I'm still very new to SASS and no programming guru.
I have ten asides elements that all require different background colours based on their class name.
I've looked through the SASS documentation and I can't figure it out.
I want to say if aside has a class name of x make background colour x if aside has a class name of y make background colour y etc
Is there a nice efficient way of doing this?
Thanks guys and sorry if its a simpleton question.
If you're using colors that don't have "standard" names (or the name of the class isn't going to be the name a color at all, eg. products = blue, addresses = red), a list of lists is what you want:
$colors:
( black #000
, white #FFF
, red #F00
, green #0F0
, blue #00F
);
#each $i in $colors {
aside.#{nth($i, 1)} {
background: nth($i, 2);
}
}
Output:
aside.black {
background: black; }
aside.white {
background: white; }
aside.red {
background: red; }
aside.green {
background: lime; }
aside.blue {
background: blue; }
If you're using colors with standard keywords, this could work:
$colors2: black, white, red, green, blue;
#each $i in $colors2 {
aside.#{$i} { background: $i; }
}
Output (though this only seems to work with --style debug, using --style compress generates errors.. weird):
aside.black {
background: black; }
aside.white {
background: white; }
aside.red {
background: red; }
aside.green {
background: green; }
aside.blue {
background: blue; }
This is simply down to how much typing you want to do. You could make a background mixin and include it within the aside CSS rule, but is that really necessary?
If it is though....
#mixin bg($color) {
background: $color;
}
aside#foo {
#include bg(#fff);
}
"if aside has a class name of x make background colour x if aside has a class name of y make background colour y" translates to the following CSS:
aside.x {background-color: x}
aside.y {background-color: y}
Is there a reason you want to use SASS? Is it to make it dynamic so that you can add any class you want in the future without updating the CSS? (If so that's not possible with SASS because the SASS code compiles to CSS and doesn't change after).
To make that work you'd have to use JS (jQuery):
$('aside').each(function () {
$(this).css('background-color', $(this).attr('class'));
});
Edit: You could use loops in SASS to generate a large number of classes and corresponding background-colors though.

Scoping sass variable

Is there a way to add scope to sass variables?
I want to be able to attach a class to my body element. The class will refer to a set of colours that the rest of the stylesheets can access.
I have tried:
#mixin theme_one{
$color: #000;
}
.theme_one{
#include theme_one;
}
and
.theme_one{
$color: #000;
}
I've just come across the same issue myself. I wanted to have different colour themes for different sections of my site.
Using a mixin seems like the best way to go. It's nicely DRY, and easy to use. The trick is not setting your colours in your main styles blocks, but rather using only the mixin for this.
I've set up the theme colours as variables at the top so they can be edited nicely, and I've set them as lists so that multiple values can be passed without hordes of variable being defined.
So:
// Variable Definitions
$defaultColor: black white grey;
$color2: blue green brown;
$color3: red white blue;
#mixin colorSet($color: $defaultColor) {
$link: nth($color, 1);
$border: nth($color, 2);
$background: nth($color, 3);
border-color: $border;
background-color: $background;
.column {
border-color: lighten($border, 10%);
}
a {
color: $link;
&:hover {
color: darken($link, 15%);
}
}
}
// Default colours
body {
#include colorSet();
}
// Scoped colours
.my-theme-3 {
#include colorSet($color3);
}
.my-theme-2 {
#include colorSet($color2);
}
Will produce something like this:
body {
border-color: white;
background-color: grey; }
body .column {
border-color: white; }
body a {
color: black; }
body a:hover {
color: black; }
.my-theme-3 {
border-color: white;
background-color: blue; }
.my-theme-3 .column {
border-color: white; }
.my-theme-3 a {
color: red; }
.my-theme-3 a:hover {
color: #b30000; }
.my-theme-2 {
border-color: green;
background-color: brown; }
.my-theme-2 .column {
border-color: #00b300; }
.my-theme-2 a {
color: blue; }
.my-theme-2 a:hover {
color: #0000b3; }
Edit: Updated to use default mixin values.
In your case no need to use mixin, If you have set of many styles then use mixin,
ie. if you have
#mixin theme_one{
$color: #000;
height: 50px;
}
then use Mixin
otherwise for single property use only variable
$color: #fff;
.some_class01{
color: $color;
background: $color;
}
.some_class22{
border-color: $color;
}
IMP: Variable should assign at the top of your code, it means don't use it after/below where you assigned it :)
Not sure if this is what you are looking for. It looks like you may have tried something similar to this,
which should probably work. (it may just be a matter of using !default)
Your body tag with a class on it..
<body class="theme_one">
</body>
Sass variables defined in stylesheet..
//THEME ONE VARIABLES
.theme_one{
$borderColor:#333 !default;
$fontColor:#999 !default;
}
//THEME TWO VARIABLES
.theme_two{
$borderColor:#CCC !default;
$fontColor:#000 !default;
}
Pre-existing CSS which will be overwritten depending on which class is used on the body tag.
h1.someheader {
color:$fontColor;
border-bottom:1px solid;
border-color:$borderColor;
}
Otherwise you could maybe try something like this. It looks like you may have tried something similar, however there seems to be an error with your mixin ... see note below.
//mixin used to set variables for properties
#mixin themeOne($fontColor,$borderColor) {
color:$fontColor;
border-color:$borderColor;
}
#include themeOne(#000,#CCC);
Pre-existing CSS
h1.someheader {
color:$fontColor
border-color:$borderColor;
border-bottom:1px solid;
}
Also note in your mixin example you are using $color:#000; ... This won't be interpreited properly as it should be color:#000; You can't use variables as selectors
unless you do something like #{$color}:#000;
I haven't quite tested this yet, so some things might need to be adjusted. If this doesn't solve your problem I hope it at least gives you some ideas.

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